Coretractable Modules

J. Aust. Math. Soc. 86 (2009), 289–304 doi:10.1017/S1446788708000360

CORETRACTABLE MODULES

B. AMINI ˛, M. ERSHAD and H. SHARIF

(Received 18 February 2007; accepted 1 November 2007)

Communicated by J. Du

Abstract

An R-module M is called coretractable if HomR(M/K, M) 6= 0 for any proper submodule K of M. In this paper we study coretractable modules and their endomorphism rings. It turns out that if all right R-modules are coretractable, then R is a right Kasch and (two-sided) perfect ring. However, the converse holds for commutative rings. Also, for a semi-injective coretractable module MR with S = EndR(M), we show that u.dim(S S) = corank(MR).

2000 Mathematics subject classiﬁcation: primary 16D10; secondary 16D40, 16L30. Keywords and phrases: coretractable module, Kasch ring, perfect ring, semi-Artinian ring, max ring, semi-injective module.

1. Introduction Throughout this paper, all rings are associative with identity and all modules are unitary right modules unless stated otherwise. Homomorphisms between modules are written and composed on the opposite side of scalars. Semisimple rings are in the sense of Wedderburn and Artin: these are rings which are semisimple as right (or left) modules over themselves. By a regular ring, we mean a ring R such that x ∈ x Rx for any x ∈ R (that is, a von Neumann regular ring). For a module MR we write Soc(MR), Rad(MR) and E(MR) for the socle, the Jacobson radical and the injective hull of MR, respectively. Also J(R) will be used for the Jacobson radical of a ring R. The notation K ≤ M, K M and K ¢ M will denote that K is a submodule, a superﬂuous submodule and an essential submodule of M, respectively. Let S be a ring and let S M be a left S-module. Then for any X ⊆ M and Y ⊆ S, the left annihilator of X in S and the right annihilator of Y in M are

lS(X) = {s ∈ S | sx = 0 for all x ∈ X}

c 2009 Australian Mathematical Society 1446-7887/2009 $16.00

289

Downloaded from https://www.cambridge.org/core. IP address: 170.106.202.58, on 01 Oct 2021 at 17:56:23, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788708000360 290 B. Amini, M. Ershad and H. Sharif [2]

and rM (Y ) = {m ∈ M | ym = 0 for all y ∈ Y },

respectively. Also for a right R-module NR, rR(Z) and lN (W) are defined similarly for any Z ⊆ N and W ⊆ R. Following Khuri [5], an R-module M is said to be retractable if HomR(M, N) 6= 0 for any nonzero submodule N of M. Retractable modules have been investigated by some authors (see, for example, [4–6, 9, 13, 14]). Dually an R-module M is called coretractable if HomR(M/K, M) 6= 0 for any proper submodule K of M. In Section 2 we study some properties of coretractable modules. We show that coretractability is preserved under finite direct sums. Also we observe that a ring R is right Kasch if and only if RR is coretractable, but for modules, coretractability and being Kasch are not equivalent conditions. In Section 3, we call a ring R right CC (completely coretractable) if all right R-modules are coretractable. It is proved that right CC rings are two-sided perfect. However, an Artinian ring need not be right CC. Also we find conditions under which free modules are coretractable. Section 4 is concerned with the endomorphism rings of coretractable modules. For a coretractable module MR with S = EndR(M) we show that S is a regular ring if and only if MR is semisimple. Moreover, if MR is semi-injective, then u.dim(S S) = corank(MR).

2. Coretractable modules It is natural to consider the dual notion of retractable modules. In this section we deﬁne the concept of coretractability for modules.

DEFINITION 2.1. An R-module M is called coretractable if, for any proper submodule K of M, there exists a nonzero homomorphism f : M −→ M with f (K ) = 0, that is, HomR(M/K, M) 6= 0. One can easily see that an R-module M is coretractable if and only if HomR(M/K, M) 6= 0 for any proper essential submodule K of M. (For if L is a submodule of M, then there exists L0 ≤ M such that L ⊕ L0 ¢ M.)

EXAMPLE 2.2. (a) Clearly semisimple modules and cogenerators (in the category Mod-R) are coretractable. (b) The Prufer¨ p-group Zp∞ is a coretractable Z-module (since, for any proper ∼ submodule K of Zp∞ , Zp∞ /K = Zp∞ ). = (c) Since HomZ(Q/Z, Q) 0, the Z-module Q is not coretractable. (d) Let n > 1 be a positive integer. It is easy to check that the Z-module Zn = Z/nZ is coretractable. We now obtain some properties of coretractable modules.

PROPOSITION 2.3. Let M be a coretractable R-module. (a) If Rad(M) 6= M, then Soc(M) 6= 0. (b) If M is nonsingular, then it is semisimple.

PROOF. (a) Let K be a maximal submodule of M. Since HomR(M/K, M) 6= 0 and M/K is a simple R-module, Soc(M) 6= 0. (b) Suppose on the contrary that M is not semisimple. So M has a proper essential submodule K . Since M/K is singular and M is nonsingular, HomR(M/K, M) = 0, which is a contradiction. 2 Since over a right nonsingular ring any projective right module is nonsingular, the following result is an immediate consequence of Proposition 2.3.

COROLLARY 2.4. Let R be a right nonsingular ring. Then a projective R-module is coretractable if and only if it is semisimple. Let R be a ring and let C be a cogenerator. Then for any module M, C ⊕ M is a cogenerator and so is a coretractable module. But M need not be coretractable. Therefore, being coretractable is not preserved by taking submodules, factor modules and direct summands. However, there are some special cases, as follows.

PROPOSITION 2.5. Let M = K ⊕ L be a coretractable module. If K is a fully invariant submodule of M or K cogenerates M, then K is also coretractable. In L Q particular, if I K or I K is coretractable for some index set I , then so is K .

PROOF. Let U be a proper submodule of K . There exists 0 6= f ∈ EndR(M) with f (U ⊕ L) = 0. If K is a fully invariant submodule of M, then

0 6= f |K ∈ EndR(K ) and ( f |K )(U) = 0.

Now suppose that K cogenerates M. So there exists g ∈ HomR(M, K ) with g f 6= 0. Thus 0 6= g f |K ∈ EndR(K ) and (g f |K )(U) = 0. So in both cases K is coretractable. 2

PROPOSITION 2.6. If M1, M2,..., Mn are coretractable R-modules, then so is Ln M = i=1 Mi . PROOF. Let K be a proper submodule of M. There exists a least integer j with K + (M1 ⊕ · · · ⊕ M j ) = M. Let

L = K + (M1 ⊕ · · · ⊕ M j−1) if j > 1 and L = K if j = 1.

So L 6= M and L ∩ M j 6= M j . There exists a nonzero homomorphism

f : M j /(L ∩ M j ) −→ M j . Since K ⊆ L, there exists a natural epimorphism h : M/K −→ M/L. Hence the composition of the following homomorphisms is nonzero: M h M M j + L ∼ M j f −→ = = −→ M j ,→ M. K L L L ∩ M j Therefore, M is coretractable. 2

In general an inﬁnite direct sum of coretractable modules is not necessarily coretractable (see Proposition 3.2 and Example 3.3). However, under certain conditions, Proposition 2.6 holds for an inﬁnite direct sum. Recall that a ring R is said to be right max if every nonzero right R-module has a maximal submodule.

PROPOSITION 2.7. Let R be a right max ring and {Mα | α ∈ I } be a family of L coretractable R-modules. Then M = α∈I Mα is also coretractable. PROOF. Let K be a proper submodule of M. There exists a maximal submodule L of M with K ⊆ L. Thus for some β ∈ I , Mβ 6⊆ L. Now, as in the proof of Proposition 2.6, we have the desired nonzero homomorphism M/K −→ M. 2 Let U and M be R-modules. U is said to be M-injective if for any monomorphism f : K −→ M and any homomorphism g : K −→ U there exists a homomorphism h : M −→ U such that h f = g. If M is M-injective, M is called quasi-injective.

PROPOSITION 2.8. Let {Mα | α ∈ I } be a family of coretractable R-modules. If for L any α, β ∈ I,Mβ is Mα-injective, then M = ∈ Mα is coretractable. In particular, α I L if M is a quasi-injective coretractable R-module, then J M is coretractable for any index set J.

PROOF. Let K be a proper submodule of M. Then there exists β ∈ I such that Mβ 6⊆ K . Therefore, there exists a nonzero homomorphism

g : Mβ /(K ∩ Mβ ) −→ Mβ .

Also the natural map f : Mβ /(K ∩ Mβ ) −→ M/K

is a monomorphism. Since Mβ is Mα-injective (for any α ∈ I ), Mβ is (M/K )-injective by Anderson and Fuller [2, Proposition 16.13]. So there exists h : M/K −→ Mβ such that h f = g. Hence 0 6= ih ∈ HomR(M/K, M), where i : Mβ −→ M is the natural inclusion. 2 The following example shows that an inﬁnite direct product of coretractable modules need not be coretractable. Q EXAMPLE 2.9. Let P be the set of all prime numbers and M = p∈P Zp. Then M is not a coretractable Z-module. In fact, let = M ∈ K Zp and f HomZ(M/K, Zp) p∈P

for some p ∈ P. Then

M/K = (pM + K )/K = p(M/K ) ⊆ ker( f )

and so f = 0. Hence ∼ Y = HomZ(M/K, M) = HomZ(M/K, Zp) 0, p∈P as desired. Let M be an R-module. A submodule K is said to be dense in M if, for any y ∈ M and 0 6= x ∈ M, there exists r ∈ R such that xr 6= 0 and yr ∈ K . Obviously, any dense submodule of M is essential. The following result is in Lam [8, Proposition 8.6].

LEMMA 2.10. Let M be an R-module and K ≤ M. Then the following statements are equivalent: (a) K is a dense submodule of M; (b) HomR(M/K, E(M)) = 0; (c) HomR(P/K, M) = 0 for any submodule P with K ⊆ P ⊆ M. COROLLARY 2.11. Let M be a coretractable R-module. Then M has no proper dense submodules. The following example shows that the converse of Corollary 2.11 is not true.

EXAMPLE 2.12. Let M be as in Example 2.9. Then M is not a coretractable Z-module. However, we show that M has no proper dense submodules. Let K ¢ M and y ∈ M\K . There exists an n > 1 such that

{r ∈ Z | yr ∈ K } = nZ.

There also exists 0 6= x ∈ M such that xn = 0. Thus for any r ∈ Z with yr ∈ K , xr = 0 and hence K is not dense in M, as required. The following result gives a necessary and sufﬁcient condition for a quasi-injective or ﬁnitely generated module to be coretractable.

THEOREM 2.13. Let M be an R-module which satisﬁes one of the following conditions: (a) M is quasi-injective; (b) every proper submodule of M is contained in a maximal submodule. Then M is coretractable if and only if it has no proper dense submodules.

PROOF. The necessity is clear by Corollary 2.11. (a) Let K be a proper submodule of M. Since K is not dense in M, by Lemma 2.10, HomR(P/K, M) 6= 0 for some submodule K ⊆ P ⊆ M. So there is a nonzero homomorphism g : P −→ M with g(K ) = 0. We can extend g to h : M −→ M by hypothesis and hence HomR(M/K, M) 6= 0. (b) Let K be a proper submodule of M. There is a maximal submodule L of M with K ⊆ L. Since L is a maximal submodule of M which is not dense in M, again by Lemma 2.10, HomR(M/L, M) 6= 0 and hence HomR(M/K, M) 6= 0. 2

A ring R is said to be right Kasch if every simple right R-module can be embedded into RR (see, for example, Lam [8, p. 280]).

THEOREM 2.14. For a ring R, the following statements are equivalent: (a) R is a right Kasch ring; (b) RR is a coretractable module; (c) every ﬁnitely generated free right R-module is coretractable; (d) for any right ideal I of R, lR(I ) 6= 0; (e) RR has no proper dense submodules. PROOF. By Proposition 2.6, statements (b) and (c) are equivalent, and by Theorem 2.14, (b) and (e) are equivalent. The equivalence of (b) and (d) follows from the fact that any endomorphism of RR is given by left multiplication by some element of R. Finally, since every proper right ideal of R is contained in a maximal right ideal, one can easily see that (a) and (b) are equivalent. 2 Example 2.15 shows that being coretractable is not preserved by extensions.

EXAMPLE 2.15. Let K be a ﬁeld and let R be the ring of 2 × 2 upper triangular 0 K matrices over K . Then J = 0 0 is the Jacobson radical of R. Since J and R/J 0 K are semisimple, they are both coretractable but R is not so (for I = 0 K we have lR(I ) = 0 and now apply Theorem 2.14).

REMARK 2.16. Let M be a nonzero R-module. An R-module N is subgenerated by M if N is isomorphic to a submodule of an M-generated module. The full subcategory of Mod-R consisting of all R-modules that are subgenerated by M is denoted by σ [M] (see Wisbauer [12, p. 118]). Following Albu and Wisbauer [1], an R-module M is called a Kasch module if it contains a copy of every simple module in σ [M]. So a ring R is a right Kasch ring if and only if RR is a Kasch module. Consequently, by Theorem 2.14, RR is coretractable if and only if it is a Kasch module. In the following example we show that for an R-module M these are two different concepts.

EXAMPLE 2.17. (a) Let R = Z and let M be as in Example 2.9. Since all simple R-modules can be embedded into M, M is a Kasch module which is not coretractable. However, a Kasch module which is ﬁnitely generated or quasi-injective is coretractable. (b) Let K be a ﬁeld and let R be the ring of all matrices of the form a x y r = 0 b z  where a, b, x, y, z ∈ K. 0 0 a Let M = {r ∈ R | b = z = 0}. Then M is a right R-module with exactly two proper nonzero submodules A and B, where A = {r ∈ R | a = b = z = 0}

and B = {r ∈ R | a = b = x = z = 0}. ∼ Also M/A = B = Soc(MR) and hence MR is coretractable. Since A/B ∈ σ[M] is a simple R-module and A/B =6∼ B, we deduce that M is not a Kasch module, as required.

3. Classes of coretractable modules The aim of this section is to ﬁnd conditions under which some classes of modules are coretractable.

PROPOSITION 3.1. For a ring R, the following statements are equivalent: (a) all free R-modules are coretractable: (b) HomR(M, R) 6= 0 for every nonzero R-module M; (c) HomR(M, R) 6= 0 for every nonzero singular R-module M.

PROOF. To show that (a) implies (b), let M be a nonzero R-module. There is a free R-module F and K ≤ F such that M =∼ F/K . Since F is coretractable, ∼ HomR(F/K, F) 6= 0 and so HomR(M, R) = HomR(F/K, R) 6= 0. Clearly (b) implies (c). Finally, to prove that (c) implies (a), let F be a free R-module and K be a proper essential submodule of F. Since F/K is a (nonzero) singular R-module, HomR(F/K, R) 6= 0 and so HomR(F/K, F) 6= 0. Therefore, F is coretractable. 2 Note that by Proposition 2.7, over a right max ring all free R-modules are coretractable if R is a right Kasch ring. But by the next result we see that the condition ‘R is a right max ring’ cannot be omitted from the assertion.

PROPOSITION 3.2. Let R be a right Kasch ring which is not right max. If every right ideal of R has a maximal submodule, then there is a free R-module which is not coretractable.

PROOF. There is an R-module M with no maximal submodules. Since every right ∼ ideal of R has a maximal submodule, HomR(M, R) = 0. Let M = F/K , where F is a free R-module and K ≤ F. Then HomR(F/K, F) = 0 and so F is not coretractable. 2 The following example guarantees the existence of a ring with the required conditions of Proposition 3.2. Recall that a subset X of a ring R is said to be right (left) T -nilpotent in case for every sequence a1, a2,... in X, there is an integer n ≥ 1 such that anan−1 ··· a1 = 0 (a1a2 ··· an = 0). It is well known that for a right max ring R,J(R) is right T -nilpotent.

EXAMPLE 3.3. Let R be the ring of all N × N lower triangular matrices over a ﬁeld K which are constant on the diagonal and have only ﬁnitely many nonzero entries off the diagonal. Then R is a local ring and J = J(R) is the set of all matrices with zero diagonal. It is easy to check that J is not right T -nilpotent and so R is not a right

max ring. Let 0 6= x ∈ R have nonzero entries only in the ﬁrst column. Then x J = 0 (and thus x R =∼ R/J) which implies that R is a right Kasch ring. It is not difﬁcult to T∞ n see that n=1 J = 0. So for any nonzero proper right ideal I of R there exists n ≥ 1 such that I ⊆ J n and I 6⊆ J n+1. Since

I/(I ∩ J n+1) =∼ (I + J n+1)/J n+1 ≤ J n/J n+1,

I/(I ∩ J n+1) is an (R/J)-module and so has a maximal submodule. Therefore, I has a maximal submodule, as desired. We now consider rings over which every module is coretractable. But for convenience we give the following deﬁnition.

DEFINITION 3.4. Let R be a ring. We call R right completely coretractable (right CC) if every right R-module is coretractable. Left completely coretractable (left CC) rings are deﬁned similarly. A ring is said to be completely coretractable (CC) if it is both left and right CC. Recall that a ring R is right semi-Artinian if every right R-module has a minimal submodule (equivalently, if every cyclic right R-module has a minimal submodule; see Stenstrom [10, Proposition VIII.2.5]).

PROPOSITION 3.5. Let R be a right semi-Artinian right max ring. If R has a unique simple right R-module (up to isomorphism), then R is a right CC ring.

PROOF. Let M be a nonzero R-module and K be a proper submodule of M. There is a maximal submodule L of M with K ⊆ L. Since R is right semi-Artinian with a unique simple (right) R-module, M/L can be embedded into M and so HomR(M/K, M) 6= 0. Therefore, M is coretractable. 2 Note that a semisimple ring is (right) CC although it does not have necessarily a unique simple module.

PROPOSITION 3.6. Let R be a right max ring. If every cyclic R-module is coretractable, then R is a right CC ring.

PROOF. Let K be a proper submodule of an R-module M. There is a maximal submodule L of M containing K . Let x ∈ M\L; then M/L =∼ x R/(L ∩ x R). Thus HomR(M/L, x R) 6= 0 which implies that HomR(M/K, M) 6= 0. Therefore, M is coretractable. 2 Let R be the ring of Example 3.3. Since R is a local left perfect ring, it is right semi- Artinian with a unique simple R-module. Then as in the proof of Proposition 3.5, we can show that all ﬁnitely generated R-modules are coretractable. However, by Proposition 3.2, R is not a right CC ring. We now investigate some properties of (right) CC rings. But ﬁrst we need a lemma.

LEMMA 3.7. Let R be a right Kasch ring. If J(R) = 0, then R is semisimple.

PROOF. Let T = Soc(RR). If T 6= R, then since R is a right Kasch ring, by 2 Theorem 2.14, I = lR(T ) 6= 0. Thus (TI ) = 0. As J(R) = 0, we have TI = 0. Since R is right Kasch, T contains a copy of every simple right R-module and so I annihilates all simple right R-modules. Therefore, I ⊆ J(R) = 0 which is a contradiction. Thus T = R and hence R is semisimple. 2

PROPOSITION 3.8. Let R be a ring. If all cyclic right R-modules are coretractable, then R is a left perfect ring.

PROOF. Let J = J(R). Since R/J is coretractable as an R-module (and hence as an (R/J)-module), it is a right Kasch ring. So by Lemma 3.7, R/J is a semisimple ring. Now let M be a nonzero cyclic right R-module. Since Rad(M) 6= M, by Proposition 2.3, Soc(M) 6= 0 and hence R is right semi-Artinian. Therefore, J is left T -nilpotent (see, for example, Stenstrom [10, Corollary VIII.2.7]) and so R is a left perfect ring. 2

Let R be a ring and I be a nonzero two-sided ideal of R. Set I 0 = R and for any ordinal α > 0 we deﬁne I α = I β I if α = β + 1 for some ordinal β and α T β α I = 0≤β<α I if α is a limit ordinal. So {I | α is an ordinal} is a descending chain of ideals of R. There is an ordinal λ such that I λ = I γ for any ordinal γ ≥ λ. Now let κ(I ) = I λ.

LEMMA 3.9. Let R be a right CC ring and let J = J(R). If κ(J) = 0, then R is a right max ring.

PROOF. By Proposition 3.1, HomR(M, R) 6= 0 for any nonzero right R-module M. So to prove that R is a right max ring, it sufﬁces to show that any nonzero right ideal of R has a maximal submodule. Let I be a right ideal of R which has no maximal submodules. By Proposition 3.8, R/J is semisimple and so Corollary 15.18 of Anderson and Fuller [2] implies that I = Rad(I ) = IJ ⊆ J. It is easy to show that I ⊆ κ(J) and hence I = 0, as required. 2

THEOREM 3.10. Let R be a right CC ring and let J = J(R). Then κ(J) = 0 and so R is a (two-sided) perfect ring.

PROOF. Suppose on the contrary that L = κ(J) 6= 0. By Proposition 3.8, J is left 2 T -nilpotent and so L ⊆ JL $ L (see Anderson and Fuller [2, Lemma 28.3]). Let T = R/L. Then by Lemma 3.13, T is a right CC ring with the Jacobson radical J(T ) = J/L. Since κ(J(T )) = 0, by Lemma 3.9, T is a right max ring. Now L/L2 is a nonzero right T -module and hence has a maximal submodule. As a consequence L R also has a maximal submodule and so LJ = Rad(L R) 6= L. This is a contradiction because κ(I )I = κ(I ) for any ideal I of R. Therefore, κ(J) = 0 and by Lemma 3.9, R is a right max ring. Thus J is right T -nilpotent. Since R is a left perfect ring, it is also right perfect. 2

COROLLARY 3.11. Let R be a right CC ring. Then a right (left) R-module is Noetherian if and only if it is Artinian.

PROOF. Since R is a (two-sided) perfect ring, it is a (two-sided) semi-Artinian max ring. However, over a semi-Artinian max ring a module is Noetherian if and only if it is Artinian (see Stenstrom [10, Proposition VIII.2.1]). 2

COROLLARY 3.12. Let R be a right semi-Artinian max ring with a unique simple right R-module. Then R is a (two-sided) perfect ring. R is also a left CC ring.

PROOF. By Proposition 3.5, R is a right CC ring and so it is (two-sided) perfect. Let J = J(R). Since R has a unique simple right R-module, R/J is a simple Artinian ring and hence R has also a unique simple left R-module. As R is a left semi-Artinian max ring, again by Proposition 3.5, R is a left CC ring. 2 The converse of Theorem 3.10 holds for commutative rings. First we need a lemma.

LEMMA 3.13. Let R and S be rings and let I be an ideal of R. (a) If R is a right CC ring, then so is R/I. (b) We have R × S is right CC if and only if R and S are both right CC.

PROOF. (a) Follows from the fact that every (R/I )-module is obviously an R-module. (b) The necessity is clear by part (a). For the sufﬁciency note that every (R × S)- module is a direct sum of an R-module and an S-module. 2

THEOREM 3.14. Let R be a commutative ring. Then the following statements are equivalent: (a) R is a CC ring; (b) every cyclic R-module is coretractable; (c) R is a perfect ring.

PROOF. Clearly (a) implies (b) and, by Proposition 3.8, (b) implies (c). Now suppose that (c) holds. By Lam [7, Theorem 23.24], R = R1 × · · · × Rn where each Ri is a local perfect ring (1 ≤ i ≤ n). Since a commutative local perfect ring is a semi- Artinian max ring with a unique simple module, by Proposition 3.5, each Ri is a CC ring. Now Lemma 3.13 implies that R is a CC ring and therefore (a) is satisﬁed. 2 Note that Theorem 3.14 is not valid for noncommutative rings, as is illustrated in the following example.

EXAMPLE 3.15. Let K be a division ring and let R be the ring of all 4 × 4 matrices of the form

a x 0 0 0 b 0 0   0 0 b y 0 0 0 a

where a, b, x, y ∈ K . Then R is a quasi-Frobenius (and hence two-sided perfect and Kasch) ring and J = J(R) consists of all matrices in R with zero diagonal. Let

1 0 0 0 0 0 0 0 e =   ∈ R. 0 0 0 0 0 0 0 1

Then Soc(eR) = eJ is a simple R-module. Also eR/eJ is a simple R-module which is not isomorphic to eJ. Therefore, HomR(eR/eJ, eR) = 0 and so eR is not coretractable. In particular, R is not a right CC ring. For the details see Lam [8, Example 16.19(4)]. Note that in Example 3.15 since R is a Kasch and max ring, all free R-modules are coretractable. But the projective module eR is not coretractable.

4. Endomorphism ring of coretractable modules

Throughout this section, MR is a right R-module and S = EndR(M) is the ring of R-endomorphisms of M. Hence S MR is a bimodule. We begin with some characterizations of coretractable modules in terms of their endomorphism rings.

LEMMA 4.1. For a right R-module M, the following statements are equivalent:

(a) MR is coretractable; (b) for any proper submodule K of MR, rM (lS(K ))/K M/K; (c) for any submodule K of MR, if rM (lS(K )) is a summand of MR, then K = rM (lS(K )).

PROOF. To prove that (a) implies (b), let K be a proper submodule of MR and L = rM (lS(K )). Suppose that

(L/K ) + (U/K ) = M/K.

If U 6= M, then by hypothesis, there exists 0 6= f ∈ S with f (U) = 0. Since K ⊆ U, f ∈ lS(K ) and so f (L) = 0. Hence f (M) = f (L + U) = 0, which is a contradiction. Therefore, L/K M/K . We now show that (b) implies (c). Let K be a proper submodule of MR and let L = rM (lS(K )) be a summand of MR. So M = L ⊕ U for some U ≤ MR. Thus

(L/K ) + ((U + K )/K ) = M/K.

Since L/K M/K , U + K = M which implies that K = L, as desired. Finally, to check that (c) implies (a), let K ≤ MR and HomR(M/K, M) = 0. Then lS(K ) = 0 and so rM (lS(K )) = M is a summand of M. Therefore, K = M and hence MR is coretractable. 2

Recall that an R-module MR is co-semisimple if for any K ≤ MR, Rad(M/K ) = 0 (see Wisbauer [12, p. 190]).

COROLLARY 4.2. Let MR be a coretractable module. Then: (a) for any K MR, we have rM (lS(K )) MR; (b) if K = Rad(MR) or K is a maximal submodule of MR, then rM (ls(K )) = K (that is, M cogenerates M/K ); (c) if MR is co-semisimple, then rM (lS(K )) = K for any K ≤ MR.

PROOF. (a) This follows from the fact that for K ≤ L ≤ M if K M and L/K M/K , then L M. (b) and (c) In both cases, since Rad(M/K ) = 0, M/K has no nonzero superﬂuous submodules. 2 The following result shows that when the endomorphism ring of a coretractable module is coretractable.

PROPOSITION 4.3. Let MR be coretractable. Then SS is coretractable if and only if IM 6= M for any proper right ideal I of S.

PROOF. In general, if SS is coretractable, then by Theorem 2.14, for any proper right ideal I of S, there exists 0 6= f ∈ S with f I = 0. Thus f (IM) = 0, which implies that IM 6= M. Conversely, let I be a proper right ideal of S. Since IM 6= M, there exists 0 6= f ∈ S with f (IM) = 0. Therefore, f I = 0 and, again by Theorem 2.14, SS is coretractable. 2 Note that the endomorphism ring of a coretractable module need not be coretractable. For example, let R = Z and M = Zp∞ . Then S = EndR(M) is a domain which is not a coretractable S-module by Theorem 2.14. We know that the endomorphism ring of a semisimple module is a regular ring. For coretractable modules the converse does also hold.

PROPOSITION 4.4. Let MR be coretractable. If S is a regular ring, then MR is semisimple.

PROOF. Suppose on the contrary that MR is not semisimple. So MR has a proper essential submodule K . Then f (K ) = 0 for some 0 6= f ∈ S. There exists g ∈ S such 2 that f = f g f . Since (g f ) = g f and ker( f ) = ker(g f ), ker( f ) is a summand of MR which contains the essential submodule K , and this is a contradiction. 2

Following Wisbauer [12, p. 261], an R-module MR is called semi-injective if for any f ∈ S, S f = lS(ker( f )) = lS(rM ( f )) (equivalently, for any monomorphism f : N −→ M, where N is a factor module of MR, and for any homomorphism g : N −→ M, there exists h : M −→ M such that h f = g). Clearly every quasi-injective module is semi-injective. Also for a ring R, RR is semi-injective if and only if it is right principally injective. In the

following proposition, we establish relations between submodules of a coretractable module and its endomorphism ring.

PROPOSITION 4.5. Let MR be a coretractable module and let K ≤ MR and I ≤ S S. (a) If lS(K ) ¢ S S, then K MR. (b) If I ¢ S S, then rM (I ) MR.

Moreover, suppose that MR is semi-injective. (c) If K MR, then lS(K ) ¢ S S. (d) If K is a maximal submodule of MR, then lS(K ) is a minimal left ideal of S. (e) If I is a minimal left ideal of S, then rM (I ) is a maximal submodule of MR.

PROOF. (a) Let L be a proper submodule of MR. There exists 0 6= f ∈ S with f (L) = 0. Since lS(K ) ¢ S S, there exists g ∈ S such that 0 6= g f ∈ lS(K ). Therefore, g f (K + L) = 0 and hence K + L 6= M. Consequently, K MR. (b) This follows from part (a) and the fact that I ⊆ lS(rM (I )). (c) Suppose that S f ∩ lS(K ) = 0 for some f ∈ S. Since MR is semi-injective,

0 = S f ∩ lS(K ) = lS(ker( f )) ∩ lS(K ) = lS(ker( f ) + K ).

But MR is coretractable and so ker( f ) + K = M. Therefore, ker( f ) = M which implies that f = 0. Hence lS(K ) ¢ S S. (d) As M is coretractable, lS(K ) 6= 0. Let 0 6= f ∈ lS(K ). Since f (K ) = 0 and K is a maximal submodule of MR, ker( f ) = K . Thus lS(K ) = lS(ker( f )) = S f . (e) Suppose that rM (I ) ⊆ K for a proper submodule K of MR. So

0 6= lS(K ) ⊆ lS(rM (I )).

Since MR is semi-injective and I is a cyclic left ideal of S, I = lS(rM (I )). Thus lS(K ) = I and so K ⊆ rM (lS(K )) = rM (I ). Therefore, rM (I ) is a maximal submodule of MR. 2 A nonzero module is said to be uniform if each nonzero submodule is essential. It is said to be hollow (co-uniform) if each proper submodule is superﬂuous.

COROLLARY 4.6. Let MR be a nonzero coretractable module. If S S is uniform, then MR is hollow. The converse holds when MR is semi-injective.

PROOF. Suppose that S S is uniform. Let K be a proper submodule of MR. Since lS(K ) 6= 0, it is an essential left ideal of S. Now Proposition 4.5(a) implies that K MR. Therefore, MR is hollow. Now suppose that MR is semi-injective and hollow. Let 0 6= f ∈ S. Since S f = lS(rM ( f )) and rM ( f ) MR, by Proposition 4.5(c), S f ¢ S S. Consequently, S S is uniform. 2

Let R be the ring of Example 3.3. Since R is a local right Kasch ring, RR is coretractable and hollow. But R = EndR(R) is not a left uniform R-module. Thus, the semi-injectivity condition is needed in the last statement of Corollary 4.6.

COROLLARY 4.7. Let MR be a semi-injective coretractable module. Then Rad(MR) = rM (Soc(S S)). In particular, Rad(MR) = M if and only if Soc(S S) = 0.

PROOF. We have X \ lS(Rad(MR)) = lS K = lS(K ). K MR K MR

By Proposition 4.5(c), lS(K ) ¢ S S for any K MR. Thus Soc(S S) ⊆ lS(Rad(MR)), equivalently, Rad(MR) ⊆ rM (Soc(S S)). Now suppose that x ∈ M\Rad(MR). There is a maximal submodule K of MR with x 6∈ K . Let 0 6= f ∈ lS(K ). By Proposition 4.5(d), lS(K ) is a minimal left ideal of S and so f ∈ Soc(S S). Since f (x) 6= 0, x 6∈ rM (Soc(S S)). Therefore, Rad(MR) = rM (Soc(S S)). Now the last assertion follows from this equality. 2 As a consequence of Corollary 4.7, we see that for a semi-injective coretractable module MR, if Soc(S S) is a summand of S S, then Rad(MR) is a summand of MR.

COROLLARY 4.8. Let M be a coretractable R-module. Then

Z(S S) ⊆ { f ∈ S | f (M) MR}.

The equality holds when MR is semi-injective.

PROOF. Let f ∈ Z(S S). There is an essential left ideal I of S such that I f = 0. Thus f (M) ⊆ rM (I ) and by Proposition 4.5(b), rM (I ) MR. Therefore, f (M) MR. Now suppose that MR is semi-injective and f ∈ S with f (M) MR. Then again by Proposition 4.5(c),

lS( f ) = lS( f (M)) ¢ S S.

Therefore, f ∈ Z(S S). 2 Observe that by Corollary 4.8, a coretractable module with zero Jacobson radical has a left nonsingular endomorphism ring. Let M be a nonzero R-module. The uniform dimension (or Goldie dimension) of M is deﬁned as

u.dim(MR) = sup{k | M contains a direct sum of k nonzero submodules}. There are various dual notions for the uniform dimension (see, for example, [3, 11]). Varadarajan [11] deﬁned the corank of a nonzero module M as

k Y corank(M) = sup k | there is an epimorphism M −→ Ni with all Ni 6= 0 . i=1

We now investigate the relation between corank(MR) and u.dim(S S) for a coretractable module MR. But ﬁrst we need a lemma.

LEMMA 4.9. Let MR be a nonzero R-module and K, L ≤ MR with K + L = M. Then lS(K ∩ L) = lS(K ) + lS(L).

PROOF. Clearly lS(K ) + lS(L) ⊆ lS(K ∩ L). Now let f ∈ lS(K ∩ L). Deﬁne g, h ∈ S as follows. For any m ∈ M, let g(m) = f (x) and h(m) = f (y) if m = x + y, where x ∈ K and y ∈ L. Since f (K ∩ L) = 0, g and h are well deﬁned. Obviously, f = g + h and g ∈ lS(L) and h ∈ lS(K ), as desired. 2

PROPOSITION 4.10. Let MR be a nonzero coretractable module. Then u.dim(S S) ≥ corank(MR).

PROOF. Suppose that f : M −→ N1 × · · · × Nn is an epimorphism, where each Ni 6= 0 (1 ≤ i ≤ n). Let πi : N1 × · · · × Nn −→ Ni be the natural projection and let T Ki = ker(πi f ) for all 1 ≤ i ≤ n. For each 1 ≤ i ≤ n, put Li = j6=i K j . Since f is an epimorphism, Ki + Li = M (1 ≤ i ≤ n). If Ii = lS(Ki ), then by induction and using P P Lemma 4.9, it is easy to check that lS(Li ) = j6=i lS(K j ) = j6=i I j . Since MR is coretractable, Ii 6= 0 for each i and X 0 = lS(M) = lS(Ki + Li ) = lS(Ki ) ∩ lS(Li ) = Ii ∩ I j . j6=i

Therefore, {I1,..., In} is an independent set of nonzero left ideals of S. Thus u.dim(S S) ≥ n and hence, by deﬁnition, u.dim(S S) ≥ corank(MR). 2

PROPOSITION 4.11. Let MR be a nonzero semi-injective coretractable module. Then corank(MR) = u.dim(S S).

PROOF. Suppose that 0 6= fi ∈ S for 1 ≤ i ≤ n and {S f1, S f2,..., S fn} is an independent set of left ideals of S. Let Ki = ker( fi )(1 ≤ i ≤ n), then by semi- injectivity of MR, S fi = lS(Ki ). Thus for any i 6= j,

0 = S fi ∩ S f j = lS(Ki ) ∩ lS(K j ) = lS(Ki + K j ).

Since MR is coretractable, Ki + K j = M. By Lemma 4.9,

lS(Ki ∩ K j ) = lS(Ki ) + lS(K j ) for any i 6= j. Now let i, j, k be three distinct positive integers less than n + 1. Then

lS(Ki + (K j ∩ Kk)) = lS(Ki ) ∩ lS(K j ∩ Kk) = lS(Ki ) ∩ (lS(K j ) + lS(Kk)) = S fi ∩ (S f j + S fk) = 0.

Therefore, Ki + (K j ∩ Kk) = M. Now we can show by induction that for any T 1 ≤ i ≤ n, Ki + ( j6=i K j ) = M. Hence the natural map

ϕ : M −→ (M/K1) × · · · × (M/Kn)

is an epimorphism. Consequently, corank(MR) ≥ u.dim(S S) and by Proposition 4.10, the equality holds. 2

Acknowledgements The authors would like to thank the referee for a careful reading of the manuscript. They would also like to express their gratitude to Professor Wisbauer for providing a copy of the paper ‘Kasch modules’.

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B. AMINI, Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71454, Iran e-mail: [email protected]

M. ERSHAD, Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71454, Iran e-mail: [email protected]

H. SHARIF, Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71454, Iran e-mail: [email protected]